Suppose we have a closed (compact, without boundary) manifold $M$. Let's assume that it is orientable, although it might play no role in the question.
Now, De Rham cohomology measures how far a closed form is from being exact. I want to pose an "approximate" form of this question. For this, endow our manifold with a metric (tensor) $g$. Thus, we get the various $L^p$ norms on the space of differential forms, via the Hodge star operator.
Suppose now that we have that $H^1_{DR}(M)=0$ and that we have an $1$-form $\alpha$ which is almost closed, i.e. it can be assumed that its norm (say $L^2$ or $L^\infty$) is arbitrarily small. Is it true that then it is almost exact, i.e. we can find an exact form arbitrarily close to $\alpha$?
So far, I tried playing a little bit with the little Hodge Theory I know but I couldn't show something. It seems to me that the closedness of the $d$ operator could be useful, but I also couldn't make that work.
I have no clue of relevant literature or how to even approach this question, so any help or ideas are much appreciated.